Class 9 SELINA Solutions Maths Chapter 24 - Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]

Ex. 24

Test Yourself

Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] Exercise Ex. 24

Solution 1(a)

Correct option: (ii) 45^o

ABC is an isosceles right-angled triangle.

⇒ AB = BC

⇒ ∠BAC = ∠BCA (angles opposite to equal sides are equal)

Given, ∠ABC = 90^o

So, in isosceles right-angled ΔABC,

∠ABC + ∠BAC + ∠BCA = 180^o

⇒ 90^o + 2∠BAC = 180^o

⇒ 2∠BAC = 90^o

⇒ ∠BAC = 45^o

Solution 1(b)

Correct option: (iii)

Given, ∠ACB = 90^o and ∠CAB = 60^o

So, in right-angled ΔACB,

∠ABC + ∠CAB + ∠ACB = 180^o

⇒ ∠ABC + 60^o + 90^o = 180^o

⇒ ∠ABC = 30^o

Therefore,

cot ∠ABC = cot 30^o =

Solution 1(c)

Correct option: (iii) 8 m

Solution 1(d)

Correct option: (i) 7.32 m

Now,

Solution 1(e)

Correct option: (iv) 40 cm

AC = CD (given)

⇒ ∠ADC = ∠DAC (angles opposite to equal sides are equal)

Let ∠ADC = ∠DAC = x

In right-angled ΔABC,

∠ABC + ∠ACB + ∠BAC = 180^o

⇒ 90^o + 60^o + ∠BAC = 180^o

⇒ ∠BAC = 30^o

In right-angled ΔABD,

∠ABD + ∠BAD + ∠ADB = 180^o

⇒ 90^o + x + 30^o + x = 180^o

⇒ 2x = 60^o

⇒ x = 30^o = ∠ADB

Now,

Solution 2

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

Solution 3

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

Solution 4

The figure is drawn as follows:

The above figure we have

Again

Now

Solution 5

(i)

From the right triangle ABE

Therefore AE = BE = 50 m.

Now from the rectangle BCDE we have

DE = BC = 10 m.

Therefore the length of AD will be:

AD = AE + DE = 50 + 10 = 60 m.

(ii)

From the triangle ABD we have

Solution 6

$F r o m space r i g h t space t r i a n g l e space A B C comma tan 60 degree equals fraction numerator A C over denominator B C end fraction rightwards double arrow square root of 3 equals fraction numerator A C over denominator 40 end fraction rightwards double arrow A C equals 40 square root of 3 space c m F r o m space r i g h t space t r i a n g l e space B D C comma tan 45 degree equals fraction numerator D C over denominator B C end fraction rightwards double arrow 1 equals fraction numerator D C over denominator 40 end fraction rightwards double arrow D C equals 40 space c m F r o m space t h e space f i g u r e comma space i t space i s space c l e a r space t h a t space A D equals A C minus D C rightwards double arrow A D equals 40 square root of 3 minus 40 rightwards double arrow A D equals 40 open parentheses square root of 3 minus 1 close parentheses rightwards double arrow A D equals 29.28 space c m$

Solution 7

We know, diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex.

The figure is shown below:

Now

And

Also given

In right triangle AOB

Also

Therefore,

Solution 8

Consider the figure

From right triangle ACF

From triangle DEB

Given , So

Therefore

Thus AB = AC + CD + BD = 54.64 cm.

Solution 9

First draw two perpendiculars to AB from the point D and C respectively. Since AB|| CD therefore PMCD will be a rectangle.

Consider the figure,

(i)

From right triangle ADP we have

Similarly from the right triangle BMC we have BM = 10 cm.

Now from the rectangle PMCD we have CD = PM = 20 cm.

Therefore

AB = AP + PM + MB = 10 + 20 + 10 = 40 cm.

(ii)

Again from the right triangle APD we have

Therefore the distance between AB and CD is .

Solution 10

From right triangle AQP

Also from triangle PBR

Therefore,

AB = AP + PB =.

Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] Exercise Test Yourself